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To understand the definition (from Wikipedia) enter image description here

I have taken a function $$ f(x) = \begin{cases} 5 \quad &\text{if $x \le 2,$} \\ 5 \quad &\text{if $x=3$,} \\ 5 \quad &\text{if $x\ge 4$,} \end{cases} $$

I also see that $\forall \epsilon \gt 0, \exists \delta=2 \gt 0, \forall x \in D, 0 \lt |x-3| \lt \delta \implies |f(x) - 5| \lt \epsilon$ where $D = \mathbb{R} \setminus \{(2,3) \cup (3,4)\}$

With this, it seems the limit for $f(x)$ exists at $x=3$ even tough there is a big gap around $x=3$. Wondering if I'm misinterpreting the definition! Is there any bound on the gap in the neighborhood of $x=c$, that I'm not able to gather from the definition of limit.

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    $\begingroup$ Why is $f$ discontinuous? $\endgroup$ Sep 3, 2020 at 18:03

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It seems like you're quoting this passage from Wikipedia. Note the condition stated at the beginning of that paragraph, that $c$ has to be a limit point of $D$. If it's not (like $c=3$ in your example), then it's not meaningful to even begin talking about “$\lim\limits_{x \to c} f(x)$”, so the limit doesn't exist in that case.

By the way, your function is actually continuous (since, well, it satisfies the definition of continuity, as can be verified). Its domain is disconnected, but that's a different concept.

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